Problem: The lifespans of snakes in a particular zoo are normally distributed. The average snake lives $29.3$ years; the standard deviation is $5.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a snake living less than $35.2$ years.
Solution: $29.3$ $23.4$ $35.2$ $17.5$ $41.1$ $11.6$ $47$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $29.3$ years. We know the standard deviation is $5.9$ years, so one standard deviation below the mean is $23.4$ years and one standard deviation above the mean is $35.2$ years. Two standard deviations below the mean is $17.5$ years and two standard deviations above the mean is $41.1$ years. Three standard deviations below the mean is $11.6$ years and three standard deviations above the mean is $47$ years. We are interested in the probability of a snake living less than $35.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the snakes will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the snakes will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $23.4$ years and the other half $({16\%})$ will live longer than $35.2$ years. The probability of a particular snake living less than $35.2$ years is ${68\%} + {16\%}$, or $84\%$.